Một máy nhiệt hoạt động theo chu trình Carnot thuận nghịch với 2 nguồn có nhiệt độ 627^circ C và -173^circ C . Nếu nó nhả một lượng nhiệt bằng 1 kJ cho nguồn lạnh trong mỗi chu trình thì công mà nó sinh ra trong mỗi chu trình là: A. 7kJ. B. 10 kJ. C C. 8kJ D. 9kJ. Clear my choice

<p> The problem involves an application of the Carnot Cycle and the principles of Thermodynamics. First, we must convert the given Celsius temperatures to the Kelvin scale, because all thermodynamic processes use the absolute Kelvin temperatures for calculation. The higher temperature Th=627°C = 627+273 = 900K. Similarly, The lower temperature Tc=-173°C = -173+273=100K.<br />The Carnot cycle operates between these two temperatures, absorbing heat from the hot source and transferring some of it to the cold source. It's given that the amount of heat transferred to the cold source in each cycle is 1kJ.<br />Let Q₂ represent the heat absorbed from the hot reservoir. The minimum possible efficiency of any heat engine is given by (Th-Tc)/Th. The engine can't deliver 100% work of the heat intake because of energy conversion's second law. Hence the efficiency is also equal to the Net work done / heat absorbed = W / Q₂.<br />Using these two expressions for efficiency and making a bit formation will get the output work W=(Q₂-Absorbed heat/Q₂)*Efficiency.<br />Substitute Q₂ = 1kJ from the givens and the fact that higher efficiency is likely will lead us on nearly choosing correct option from the multiple choices.</p>